Really hard one! (Instead of $I_1,I_2,I_3,I_4$ i use $X,Y,Z,W$) We are going to use the (well-known) lemma Consider a triangle $ABC$ and a point $P$ on its side $AC$. Let the incircles of triangles $PAB$ and $PBC$ have centers $X$ and $Y$, respectively; the incircle of triangle $ABC$ touches $AC$ at $U$ .Then the points $P$ and $U$ lie on the circle with diameter $XY$ Also, it is well known, as $ABCD$ is tangential, that the incircles centered at $J_4$ and $J_2$ will touch $BD$ at a common point (say $P$) and similarly the incircles centered at $J_3$ and $J_1$ share a common point on $AC$ (say $U$) Let $R$ be the excimilicenter of $(J_4), (J_2)$. Let O be the center of the incircle of $ABCD$. Now, the exsimilicenter of $(O)$ and $(J_4)$ is $A$, and the exsimilicenter of $(O)$ and $(J_2)$ is $C$. Hence, the $1$st Monge theorem yields that $R$ lies on the line $AC$ The lines $AX$ and $CY$ meet at $J_1$; the lines $XJ_4$ and $YJ_2$ meet at $B$; the lines $AJ_4$and $CJ_2$ meet at $O$. Since $B, O, J_1$ are collinear, we can apply the Desargues theorem to triangles $AXJ_4$ and $CYJ_2$, and infer that the lines $AC, XY$ and $J_4J_2$ concur. Since $AC$ and $J_4J_2$ meet at $R$, we see that $R$ lies on $XY$. Hence, $R$ is the exsimilicenter of $(X)$ and$ (Y)$. Similarly, $R$ is the exsimilicenter of $(Z)$ and $(W)$ and lies on the line $ZW$. Moreover, there will be a similar point $Q$ on $BD$ with analogous properties Of course, the point $P$ lies on $J_4J_2$; moreover, the points $P$ and $R$ divide the segment $J_4J_2$ harmonically, since $P$ is the internal center of similtude of $(J_4)$ and $(J_2)$ and$ R$ is their external center of similtude. Now, we can apply the lemma stated at the begining to triangle $ABC$ and the point $T$ on its side $AC$. Consequently, the points $T$ and $U$ lie on the circle with diameter $XY$. Hence, the points $X, Y, P, U $ are concylic, and $RX \cdot RY = RT \cdot RU$. Similarly, $RZ \cdot RW = RT \cdot RU$, and thus $RX \cdot RY = RZ \cdot RW$, and the points$ X, Y, Z, W$ are concyclic Now, regarding the center of $XYZW$...... (remember we denote by $Q$ the exsimilicenter of $(J_1), (J_3)$) $\triangle TQR$ is the diagonal triangle of the cyclic quadrilateral $XYZW$; hence, it is autopolar with respect to the circumcircle of this cyclic quadrilateral. (In fact, it is well-known that the diagonal triangle of a cyclic quadrilateral is autopolar with respect to the circumcircle.) Hence, the line $TR$ (line $AC$) is the polar of $Q$ with respect to the circle through $X, Y, Z, W$. Consequently, the lines $AC$ and $O'Q$ are perpendicular ($O'$ is the center of the circle through $X, Y, Z, W$). On the other hand, the line $J_3J_1$ is perpendicular to $AC$ , too (since the circles $(J_1)$ and $(J_3)$ touch $AC$ at one point $(U)$, and it also passes through $Q$. Hence, the line $O'Q$ coincides with the line $J_1J_3$ , $\implies $ the point $O'$ lies on $J_1J_3$. Similarly, $O'$ lies on $J_4J_2$. So $O'= J_1J_3 \cap J_2J_4$ This completes this LONG proof

Attachments:

The problem appears here and here.

@above Indeed. It is an old problem. I used much the same of its proof when doing this a year ago. LoloChen wrote: P79: Let ABCD has both an incircle and a circumcircle. AC cross BD at P. Prove that the incenters of ABC,BCD,CDA,DAB,ABP,BCP,CDP,DAP are concyclic. (not so hard and possibly famous)